C4 January 2011 Q7
7. \[I = \int_2^5 \frac{1}{4 + \sqrt{(x - 1)}}\,\mathrm{d}x\]
(a) Given that \(y = \dfrac{1}{4 + \sqrt{(x - 1)}}\), complete the table below with values of \(y\) corresponding to \(x = 3\) and \(x = 5\). Give your values to 4 decimal places.
(2)
| \(x\) | 2 | 3 | 4 | 5 |
|---|---|---|---|---|
| \(y\) | 0.2 | 0.1745 |
(b) Use the trapezium rule, with all of the values of \(y\) in the completed table, to obtain an estimate of \(I\), giving your answer to 3 decimal places. (4)
(c) Using the substitution \(x = (u - 4)^2 + 1\), or otherwise, and integrating, find the exact value of \(I\). (8)
| Scheme | Marks |
|---|---|
| \(x = 3 \Rightarrow y = 0.1847\) awrt | B1 |
| \(x = 5 \Rightarrow y = 0.1667\) awrt or \(\tfrac{1}{6}\) | B1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(I \approx \underline{\underline{\dfrac{1}{2}}}\big[0.2 + 0.1667 + 2(0.1847 + 0.1745)\big]\) | B1 M1 A1ft |
| \(\approx 0.543\) 0.542 or 0.543 | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = 2(u - 4)\) | B1 |
| \(\displaystyle\int \frac{1}{4 + \sqrt{(x - 1)}}\,\mathrm{d}x = \int \frac{1}{u} \times 2(u - 4)\,\mathrm{d}u\) | M1 |
| \(\displaystyle = \int \left(2 - \frac{8}{u}\right)\mathrm{d}u\) | A1 |
| \(= 2u - 8\ln u\) | M1 A1 |
| \(x = 2 \Rightarrow u = 5,\quad x = 5 \Rightarrow u = 6\) | B1 |
| \(\Big[2u - 8\ln u\Big]_5^6 = (12 - 8\ln 6) - (10 - 8\ln 5)\) | M1 |
| \(= 2 + 8\ln\left(\dfrac{5}{6}\right)\) | A1 |
| (8) | |
| (14 marks) |